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(A) The amount of a radioactive substance remaining after time $t$ is given by $N = \frac{N_0}{2^{t/T_{1/2}}}$.Given: Initial ratio $\frac{N_{01}}{N_{

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$1:1$

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(A) The amount of a radioactive substance remaining after time $t$ is given by $N = \frac{N_0}{2^{t/T_{1/2}}}$.Given: Initial ratio $\frac{N_{01}}{N_{02}} = \frac{2}{1}$,$T_{1/2,1} = 12 	ext{ hours}$,$T_{1/2,2} = 16 	ext{ hours}$,and time $t = 2 	ext{ days} = 48 	ext{ hours}$.The ratio of remaining amounts is $\frac{N_1}{N_2} = \frac{N_{01} \cdot 2^{-t/T_{1/2,1}}}{N_{02} \cdot 2^{-t/T_{1/2,2}}}$.Substituting the values: $\frac{N_1}{N_2} = \left(\frac{2}{1}ight) \cdot \frac{2^{-48/12}}{2^{-48/16}} = 2 \cdot \frac{2^{-4}}{2^{-3}} = 2 \cdot 2^{-4+3} = 2 \cdot 2^{-1} = 2 \cdot \frac{1}{2} = 1$.Therefore,the ratio is $1:1$.

At a certain moment,the amounts of two radioactive substances are in the ratio $2:1$. Their half-lives are $12 \text{ hours}$ and $16 \text{ hours}$ respectively. What will be the ratio of their remaining amounts after $2 \text{ days}$?

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